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2 years ago

7. Explain why observations of comets and shooting stars presented challenges to ancient astronomers using

1 answer:

3 0

According to Ptolemy's model, he, too, believed in a geocentric Universe and that the planets and stars were perfect spheres, though Earth itself was not.

He further thought that the movements of the planets and stars must be circular since they were perfect and, if the motions were circular, then they could go on forever.

<h3>What is comets and shooting stars?</h3>

Shooting stars are very different from comets, although the two can be related. A Comet is a ball of ice and dirt, orbiting the Sun (usually millions of miles from Earth). ... A shooting star on the other hand, is a grain of dust or rock (see where this is going) that burns up as it enters the Earth's atmosphere.

Learn more about ptolemy's model:

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PLEASE NEED AN ANSWER SO I CAN SUBMIT IT!!! THANK YOU IN ADVANCE (will give brainliest)

alexandr402 [8]

Answer:

Im only 12 and i need the points so ima try my best.

Explanation:

574.780616 m6 kg3 s-6 K-3 mol-3

6 0

3 years ago

There are four charges, each with a magnitude of 4.25 C. Two are positive and two are negative. The charges are fixed to the cor

VMariaS [17]

Answer:

F = 7.68 10¹¹ N, θ = 45º

Explanation:

In this exercise we ask for the net electric force. Let's start by writing the configuration of the charges, the charges of the same sign must be on the diagonal of the cube so that the net force is directed towards the interior of the cube, see in the attached numbering and sign of the charges

The net force is

F_ {net} = F₂₁ + F₂₃ + F₂₄

bold letters indicate vectors. The easiest method to solve this exercise is by using the components of each force.

let's use trigonometry

cos 45 = F₂₄ₓ / F₂₄

sin 45 = F_{24y) / F₂₄

F₂₄ₓ = F₂₄ cos 45

F_{24y} = F₂₄ sin 45

let's do the sum on each axis

X axis

Fₓ = -F₂₁ + F₂₄ₓ

Fₓ = -F₂₁₁ + F₂₄ cos 45

Y axis

F_y = - F₂₃ + F_{24y}

F_y = -F₂₃ + F₂₄ sin 45

They indicate that the magnitude of all charges is the same, therefore

F₂₁ = F₂₃

Let's use Coulomb's law

F₂₁ = k q₁ q₂ / r₁₂²

the distance between the two charges is

r = a

F₂₁ = k q² / a²

we calculate F₂₄

F₂₄ = k q₂ q₄ / r₂₄²

the distance is

r² = a² + a²

r² = 2 a²

we substitute

F₂₄ = k q² / 2 a²

we substitute in the components of the forces

Fx = $- k \frac{q^2}{a^2} + k \frac{q^2}{2 a^2} \ cos 45$

Fx = $k \frac{q^2}{a^2}$ ( -1 + ½ cos 45)

F_y = k \frac{q^2}{a^2} ( -1 + ½ sin 45)

We calculate

F₀ = 9 10⁹ 4.25² / 0.440²

F₀ = 8.40 10¹¹ N

Fₓ = 8.40 10¹¹ (½ 0.707 - 1)

Fₓ = -5.43 10¹¹ N

remember cos 45 = sin 45

F_y = - 5.43 10¹¹ N

We can give the resultant force in two ways

a) F = Fₓ î + F_y ^j

F = -5.43 10¹¹ (i + j) N

b) In the form of module and angle.

For the module we use the Pythagorean theorem

F = $\sqrt{F_x^2 + F_y^2}$

F = 5.43 10¹¹ √2

F = 7.68 10¹¹ N

in angle is

θ = 45º

7 0

3 years ago

What is the magnitude of the momentum of a 3400 kg airplane traveling at a speed of 450 miles per hour?

Vitek1552 [10]

Answer:

The magnitude of momentum of the airplane is $6.83\times 10^5\ kg-m/s$ .

Explanation:

Given that,

Mass of the airplane, m = 3400 kg

Speed of the airplane, v = 450 miles per hour

Since, 1 mile per hour = 0.44704 m/s

v = 201.16 m/s

We need to find the magnitude of momentum of the airplane. It is given by the product of mas and velocity such that,

$p=mv$

$p=3400\ kg\times 201.16 \ m/s$

$p=683944\ kg-m/s$

$p=6.83\times 10^5\ kg-m/s$

So, the magnitude of momentum of the airplane is $6.83\times 10^5\ kg-m/s$ . Hence, this is the required solution.

7 0

3 years ago

Read 2 more answers

A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas

Radda [10]

Answer:

change in momentum, $\Delta p=7.65 \,kg.m.s^{-1}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

Average Force, $F=144.3396\,N$

$F_x=125.0018\,N$

$F_y=72.1698\,N$

Explanation:

Given:

angle of kicking from the horizon, $\theta= 30^{\circ}$

velocity of the ball after being kicked, $v=18 m.s^{-1}$

mass of the ball, $m=0.425\, kg$

time of application of force, $t=5.3\times 10^{-2}\,s$

We know, since body is starting from the rest

$\Delta p=m.v$ .....................(1)

$\Delta p=0.425\times 18$

$\Delta p=7.65 \,kg.m.s^{-1}$

Now the components:

$\Delta p_x= 7.65\times cos 30^{\circ}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

similarly

$\Delta p_y= 7.65\times sin 30^{\circ}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

also, impulse

$I=F\times t$ .........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

$F\times t=m.v$

$F\times 5.3\times 10^{-2}= 7.65$

$F=144.3396\,N$

Now, the components

$F_x=144.3396\times cos 30^{\circ}$

$F_x=125.0018\,N$

$F_y=144.3396\times sin 30^{\circ}$

$F_y=72.1698\,N$

6 0

3 years ago

Bob and John are pulling in different directions. If Bob is pulling to the right with a force of 10N, and John is pulling to the

kykrilka [37]

Answer:

A) -2N

B) Left

C) -0.5

Explanation:

A) -12 + 10

B) More force is acted on in that direction

C) Net force/Mass (-2/4)

5 0

3 years ago